## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.7 - Extreme Values and saddle Points - Exercises 14.7 - Page 843: 29

#### Answer

Local maximum of $f(\dfrac{1}{2},1)=-3-2 \ln 2$

#### Work Step by Step

Given: $f_x(x,y)=-4+\dfrac{2}{x}=0, f_y(x,y)=-1+\dfrac{1}{y}=0$ Simplify the given two equations. Critical points: $(\dfrac{1}{2},1)$ In order to solve this problem we will have to apply Second derivative test that suggests the following conditions to calculate the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$. 1. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\gt 0$ , then $f(a,b)$ is a local minimum. 2. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\lt 0$ , then $f(a,b)$ is a local maximum. 3. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \lt 0$ , then $f(a,b)$ is a not a local minimum or local maximum but a saddle point. $D(\dfrac{1}{2},1)=f_{xx}f_{yy}-f^2_{xy}=8 \implies D=8 \gt 0;f_{xx}=-8 \lt 0$ Hence, Local maximum of $f(\dfrac{1}{2},1)=-3-2 \ln 2$

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